Mathematics · 2024
JEE Main · 4 April 2024, Shift 1 · Q13
One of the points of intersection of the curves y=1+3 x-2 x^2 and y=1/x is (1/2, 2). Let the area of the region enclosed by these curves be 1/24(l √…
One of the points of intersection of the curves $\displaystyle y=1+3 x-2 x^2$ and $\displaystyle y=\frac{1}{x}$ is $\displaystyle \left(\frac{1}{2}, 2\right)$. Let the area of the region enclosed by these curves be $\displaystyle \frac{1}{24}(l \sqrt{5}+m)-n \log _{\mathrm{e}}(1+\sqrt{5})$, where $\displaystyle l, m, n \in \mathbf{N}$. Then $\displaystyle l+\mathrm{m}+\mathrm{n}$ is equal to
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle 30$
More from Area Under the Curve
- Let f:(1, ∞) → R be a function defined as f(x)=(x-1)/(x+1). Let f^i+1(x)=f(f^i(x)), i=1,2, …, 25, where f^1(x)=f(x). If g(x)+f^26(x)=0, x ∈(1, ∞),…2026
- The area (in square units) of the region bounded by the parabola y^2=4(x-2) and the line y=2 x-8, is:2024
- The area of the region A={(x, y): 4 x^2+y^2 ≤ 8. and.y^2 ≤ 4 x} is:2026
- If the area of the region {(x, y): a/x^2 ≤ y ≤ 1/x, 1 ≤ x ≤ 2,0<a<1} is ( log_e 2)-1/7 then the value of 7 a-3 is equal to:2024
- The area of the region R={(x, y): x y ≤ 27,1 ≤ y ≤ x^2} is equal to:2026
- The area of the region bounded by the curve y= max {|x|, x|x-2|}, the x -axis and the lines x=-2 and x=4 is equal to ____2025
- A line passing through the point A(-2,0), touches the parabola P: y^2=x-2 at the point B in the first quadrant. The area, of the region bounded by…2025
- Let f(α) denote the area of the region in the first quadrant bounded by x=0, x=1, y^2=x and y=|α x-5|-|1-α x|+α x^2. Then (f(0)+f(1)) is equal to2026
JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.